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S_A_V [24]
3 years ago
12

Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point to

ward. You have a compass that is free to swivel in any direction. Which way does your compass point? Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? It would point up. It would point east. It would point down. It would point west.
Physics
1 answer:
Brut [27]3 years ago
7 0

Answer:

It would point up.

Explanation:

Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.

Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.

So, the compass would point up.

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The percussion instruments (drums etc) are the richest producers
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Fuzzy electric guitars are next richest.

The strings and brass instruments are moderate producers of
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Then come the woodwinds ... clarinet, oboe, etc.

The closest to 'pure' tones of single frequency are the sounds
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3 years ago
He generation of a magnetic field by an electric current is
USPshnik [31]

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Which telescope would be better viewing a faint, distant star? Why?
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Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a
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A) Vector A

The x-component of a vector can be found by using the formula

v_x = v cos \theta

where

v is the magnitude of the vector

\theta is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its x-component is

A_x = A cos \theta_A = (50) cos 0^{\circ}=50

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ} (negative since it is below the x-axis), so the x-component is

B_x = B cos \theta_B = (120) cos (-70^{\circ})=41

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

v_y = v sin \theta

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its y-component is

A_y = A sin \theta_A = (50) sin 0^{\circ}=0

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ}, so the y-component is

B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

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How might what is happening in this image affect the nervous system?
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